# Is Cauchy's Integral Formula also valid for loops?

I have a question about the definition of Cauchy's Integral Formula:

Let $$\gamma$$ be a simpled closed positively oriented contour. If f is analytic in some simply connected domain $$D$$ containing $$\gamma$$ and $$z_0$$ is any point inside $$\gamma$$ then: $$f(z_0) = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-z_0} dz$$ Fundamentals of Complex Analysis (Pearson Education 2014)

My question is regarding the choice of contour above, is this not also true if $$\gamma$$ is a loop (see example image below) or why do we have to add that the contour needs to also be simple? The proof of the formula involves using the Deformation Invariance Theorem (deformation of contours) to deform the contour $$\gamma$$ in question to a circle around the point $$z_0$$ and we should still be able to do this step if the contour in question is a loop. Am I missing something obvious or are there maybe applications after complex analysis that constrains the formula, which the author might not notice the reader about in my textbook.

Edits:

• Missed the $$2\pi i$$ part in the formula.
• If it is not simple, what is the meaning of “positively oriented” then? Commented Jan 19, 2022 at 21:41
• Some versions of the formula have an additional accounting for the index of the loop around $z_0$. Commented Jan 19, 2022 at 21:43
• We can still have a closed contour that is positively oriented and if it was negatively orientated we would just have to change the sign of the integral and everything would work out just fine. please let me know if I'm missing something but is the orientation crucial for the formula and the distinction between "simple closed contours" and "closed contours" Commented Jan 19, 2022 at 21:49
• @copper.hat could you link me to any resources that go into details about that, would love to read up! thanks for the quick answer! Commented Jan 19, 2022 at 21:52
• @JamesBlond Theorem 5.4 in Conway's "Functions of one complex variable", 2nd Ed. Basically $\eta(\gamma; a)f(a) = {1 \over 2 \pi i} \int_\gamma {f(z) \over z-a} dz$. Commented Jan 19, 2022 at 22:01

First of all, notice that you missed a $$2 \pi i$$ factor in the formula.
Now to your question. The thing is, that if you integrate in such a loop, you can "divide" that loop into two closed curves and by additive properties of the integrals, write the integral as the sum of two other integrals. however, if one of those divisions of the loops doesn't include any singularities, then that integral reduces to $$0$$ (see Cauchy-Goursat Theorem).
So, in the end, it does not matter if $$\gamma$$ is a loop or just a simple-closed curve, because a loop could always be treated as a union of closed curves.